NAVIER-STOKES
Derivation of NavierStokes Equation Using cylindrical co-ordinates (r, Ɵ, z) Year 2012
PRAXIE
This document provides a step-by-step guide to deriving the NS equation using cylindrical co-ordinates. The steps have been collected from different documents available on the web; frankly speaking, this document just assembles them into a single file. Although the derivation is not the most detailed one can go for, it certainly helps the user to have a feel of the derivation process and proceed in a more detailed fashion if one decides to. Thank You :-)
NOTE: The derivation has been divided into five steps, with each step collected from a different pdf, the necessary pages of which have been attached herein and the required figures and terms either clouded out or highlighted in yellow. Remember, no detailed calculations have been shown in this document, only the superficial things necessary for one to derive the NS equation!!! U r highly encouraged to do the detailed derivation and share it on the web for others to learn and use. Cheers ☺
STEP I (Pgs. 3-5) Using the figures highlighting stresses in r, Ɵ and z directions, we find out the equilibrium equations in terms of normal and shear stresses.
STEP III (Pg. 9) Substituting the acceleration terms from step II in the equilibrium equations calculated in step I, we find out the r, Ɵ and z components of the momentum equation.
STEP II (Pgs 6-8) The material derivative or acceleration terms are derived in terms of cylindrical coordinates (r, Ɵ, z).
STEP IV (Pg. 10) The normal and shear stresses are shown in cylindrical coordinates; no derivation has been done over here!
STEP V (Pg. 11) Substituting the normal and shear stresses from step IV into the momentum equation derived in step III and using the continuity equation (in cylindrical coordinates) for simplification, we finally get the NavierStokes equation in r, Ɵ and z directions
Page 3 16
2. Governing Equations
σθθ +
∂σθθ δθ ∂θ τθr +
∂τθr δθ ∂θ τrθ +
δθ
τ zr +
σ rr
∂τzr δz ∂z
∂τrθ δr ∂r σrr +
∂σrr δr ∂r
τrθ τθr σθθ FIGURE 2.2. Stresses in the r and θ Directions.
of stress, strain, and displacement in cylindrical coordinates. The following sections provide a succinct review of essential topics needed for the establishment of the governing elasto-dynamic equations.
2.1 State of Stresses at a Point A three dimensional state of stress in an infinitesimal cylindrical element is shown in the following three figures. Figure 2.1 depicts such an element with direct stresses, dimensions, and directions of the cylindrical coordinate. Figure 2.2 represents the direct and shear stresses in the radial and transverse directions (r and θ), and the variation of direct and shear stresses in these two directions. Figure 2.3 shows direct and shear stresses associated with the planes perpendicular to the r and z directions, as well as their variations along these directions. In the above graphical representations the changes in direct and shear stresses are given by considering the first order infinitesimal term used in Taylor series approximation. The series approximation has been truncated after the second term. Further terms within the series representation contain terms of an infinitesimal length squared. Assuming that the second
Page 4 2. Governing Equations
z
σ zz + σrr
17
∂σ zz δz ∂z ∂τ zθ δθ ∂θ ∂τ τ zr + zr δr ∂r Fr
τ zθ + δz
σ θθ
τθr δr τθz
σrr +
∂σrr δr ∂r
δθ
r FIGURE 2.3. Stresses in the plane perpendicular to r and z direction.
order terms are very small, they can be neglected. Therefore, the change in stress across the element is considered very small.
2.2 Equilibrium Equations in Terms of Stress Utilizing Newton’s second law and the graphical representation of the state of stress, the equilibrium equations for an infinitesimal element in a cylindrical coordinates will be developed. By examining the state of stress on the element shown in section 2.1, the following equilibrium equation in the r direction is given. ¶ µ ¶ µ δθ ∂σrr ∂τ rθ δr (r + δr) δθδz + τ rθ + δθ δrδz cos σ rr + ∂r ∂θ 2 ¶µ ¶ µ δr ∂τ rz δz r+ δrδθ + Fr δrδθδz + τ rz + ∂z 2 ¶ µ δr δθ = σrr rδθδz + δτ rθ δrδz cos + τ rz r + δrδθ + 2 2 µ ¶ δθ ∂σθθ δθ + σθθ + δθ δrδz sin + σ θθ δrδz sin (2.1) ∂θ 2 2
Page 5 18
2. Governing Equations
Canceling appropriate terms from both sides of the equation and after simplifying, it yields: 1 ∂τ rθ ∂τ rz σ rr − σ θθ ∂σ rr + + + + Fr = 0 ∂r r ∂θ ∂z r
(2.2)
Similarly, the equilibrium equation for the θ direction yields: 1 ∂σ θθ ∂τ θz 2 ∂τ rθ + + + τ rθ + Fθ = 0 ∂r r ∂θ ∂z r
(2.3)
and finally, for the z direction one may write: 1 ∂τ θz ∂σ zz 1 ∂τ rz + + + τ rz + Fz = 0 ∂r r ∂θ ∂z r
(2.4)
In the above simplifications, due to very small angle of δθ, the following approximations were used: cos
δθ ≈1 2
sin
δθ δθ ≈ 2 2
(2.5)
In addition to the stresses, body forces acting throughout the element have been considered for each direction. These are denoted by Fr , Fθ , and Fz which are introduced as forces in the r, θ, and z direction per unit of volume. Due to the cancellation of the moments about each of the three perpendicular axes, the relations among the six shear stress components are presented by the following three equations: τ rθ = τ θr
τ θz = τ zθ
τ zr = τ rz
(2.6)
Therefore, the stress at any point in the cylinder may be accurately described by three direct stresses and three shear stresses.
2.3 Stress-Strains Relationships The constitutive relation between stresses and strains for a homogeneous and isotropic material can be expressed by Hooke’s law. By definition, a homogeneous and isotropic material has the same properties in all directions. From this, the following three equations for direct strain in terms of stress are presented: err E eθθ E ezz E
= σ rr − ν (σ θθ + σ zz ) = σ θθ − ν (σzz + σ θθ ) = σ zz − ν (σ rr + σ θθ )
(2.7) (2.8) (2.9)
Page 6
2. The Material Derivative in Cylindrical Coordinates This one a little bit more involved than the Cartesian derivation. The reason for this is that the unit vectors in cylindrical coordinates change direction when the particle is moving. In the Lagrangian reference, the velocity is only a function of time. When we switch to the Eulerian reference, the velocity becomes a function of position, which, implicitly, is a function of time as well as viewed from the Eulerian reference. Then
(Eq. 1) and the material derivative is written as (with the capital D symbol to distinguish it from the total and partial derivatives)
(Eq. 2) Special attention must be made in evaluating the time derivative in Eq. 2. In dynamics, when differentiating the velocity vector in cylindrical coordinates, the unit vectors must also be differentiated with respect to time. In this case, the partial derivative is computed at a fixed position and therefore, the unit vectors are "fixed" in time and their time derivatives are identically zero. Then, we have
(Eq. 3) we can now evaluate the remaining terms in Eq. 2 as follows
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(Eq. 4) and
(Eq. 5) finally
Page 8
(Eq. 6) When these are put together, the material derivative in cylindrical coordinates becomes
Page 9
5.7 Basic Equations in different Coordinate Systems
–
143
x2 −Component: ρ
∂U2 ∂U2 ∂U2 ∂U2 ∂P + U2 + U2 + U3 =− ∂t ∂x1 ∂x2 ∂x3 ∂x2 2 ∂ U2 ∂ 2 U2 ∂ 2 U2 +µ + + + ρg2 ∂x21 ∂x22 ∂x23 (5.110)
x3 −Component: ρ
∂U3 ∂U3 ∂U3 ∂U3 ∂P + U1 + U2 + U3 =− ∂t ∂x1 ∂x2 ∂x3 ∂x3 2 ∂ U3 ∂ 2 U3 ∂ 2 U3 +µ + + + ρg3 ∂x21 ∂x22 ∂x23 (5.111)
Momentum Equations in Cylindrical Coordinates - Momentum equations with τij −terms: ! Uϕ2 ∂Ur ∂Ur Uϕ ∂Ur ∂Ur ∂P + Ur + − + Uz =− ∂t ∂r r ∂ϕ r ∂z ∂r 1 ∂ 1 ∂τrϕ τϕϕ ∂τrz − (rτrr ) + − + + ρgr r ∂r r ∂ϕ r ∂z (5.112)
r−Component: ρ
ϕ−Component: ρ
z−Component: ρ
-
∂Uϕ ∂Uϕ Uϕ ∂Uϕ Ur Uϕ ∂Uϕ + Ur + + + Uz ∂t ∂r r ∂ϕ r ∂z 1 ∂P 1 ∂ 2 1 ∂τϕϕ ∂τϕz =− − (r τ ) + + + ρgϕ rϕ r ∂ϕ r2 ∂r r ∂ϕ ∂z (5.113)
∂Uz ∂Uz Uϕ ∂Uz ∂Uz ∂P + Ur + + Uz =− ∂t ∂r r ∂ϕ ∂z ∂z 1 ∂ 1 ∂τϕz ∂τzz − (rτrz ) + + + ρgz r ∂r r ∂ϕ ∂z (5.114)
- Navier-Stokes equations for ρ and µ equally constant:
! Uϕ2 ∂Ur ∂Ur Uϕ ∂Ur ∂Ur r−Component: ρ + Ur + − + Uz (5.115) ∂t ∂r r ∂ϕ r ∂z ∂ 1 ∂ 1 ∂ 2 Ur 2 ∂Uϕ ∂ 2 Ur ∂P =− +µ (rUr ) + 2 − + + ρgr ∂r ∂r r ∂r r ∂ϕ2 r2 ∂ϕ ∂z 2
Page 10 26
FLUID MECHANICS 1 ∂v x ∂vz + + 2 ∂z ∂x
!2
!2 1 ∂vy ∂vz + + . 2 ∂z ∂y
(2.141)
In the above, γ, µ, κ, and M are treated as uniform constants.
2.19 Fluid Equations in Cylindrical Coordinates Let us adopt the cylindrical coordinate system, r, θ, z. Making use of the results quoted in Section C.3, the components of the stress tensor are σrr
=
σθθ
=
σzz
=
σrθ = σθr
=
σrz = σzr
=
σθz = σzθ
=
∂vr , ∂r ! 1 ∂vθ vr −p + 2 µ + , r ∂θ r −p + 2 µ
∂vz , ∂z ! 1 ∂vr ∂vθ vθ µ + − , r ∂θ ∂r r ! ∂vr ∂vz µ + , ∂z ∂r ! 1 ∂vz ∂vθ µ + , r ∂θ ∂z
−p + 2 µ
(2.142) (2.143) (2.144) (2.145) (2.146) (2.147)
whereas the equations of compressible fluid flow become Dρ Dt Dvr vθ2 − Dt r
Dvθ vr vθ + Dt r
=
−ρ ∆,
=
−
=
1 ∂p ∂Ψ − ρ ∂r ∂r
+
! µ 2 vr 2 ∂vθ 1 ∂∆ ∇ vr − 2 − 2 + , ρ r r ∂θ 3 ∂r
−
1 ∂p 1 ∂Ψ − ρ r ∂θ r ∂θ
! µ 2 2 ∂vr vθ 1 ∂∆ ∇ vθ + 2 − 2+ , ρ 3r ∂θ r ∂θ r ! 1 ∂p ∂Ψ µ 2 1 ∂∆ − − + ∇ vz + , ρ ∂z ∂z ρ 3 ∂z ! κM 2 p χ+ ∇ , R ρ +
Dvz = Dt ! 1 Dρ γ p Dρ − = γ − 1 Dt ρ Dt
(2.148)
(2.149)
(2.150) (2.151) (2.152)
where ∆ = D Dt
=
∇2
=
1 ∂(r vr ) 1 ∂vθ ∂vz + + , r ∂r r ∂θ ∂z ∂ ∂ vθ ∂ ∂ + vr + + vz , ∂t ∂r r ∂θ ∂z ! 2 1 ∂ ∂ 1 ∂ ∂2 r + 2 2 + 2, r ∂r ∂r r ∂θ ∂z
(2.153) (2.154) (2.155)
Page 11 Navier - Stokes equation:
We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity r μ =const), with a velocity field V = (u ( x,y,z) , v( x,y,z) , w ( x,y,z)) Incompressible continuity equation: ∂u ∂v ∂w + + =0 eq1. ∂x ∂y ∂z
Navier - Stokes equation: r vector form:
ρ
r r DV = −∇ P + ρ g + μ ∇ 2 V Dt
x component: ⎛ ∂u ∂u ∂u ∂u ⎞ ∂P ∂ 2u ∂ 2u ∂ 2u ρ ⎜⎜ + u + v + w ⎟⎟ = − + ρg x + μ ( 2 + 2 + 2 ) ∂x ∂y ∂z ⎠ ∂x ∂x ∂y ∂z ⎝ ∂t y component: ⎛ ∂v ∂v ∂v ∂v ⎞ ∂P ∂ 2v ∂ 2v ∂ 2v ρ ⎜⎜ + u + v + w ⎟⎟ = − + ρg y + μ ( 2 + 2 + 2 ) ∂x ∂y ∂z ⎠ ∂y ∂x ∂y ∂z ⎝ ∂t z component: ⎛ ∂w ∂w ∂w ∂w ⎞ ∂P ∂2w ∂2w ∂2w ρ ⎜⎜ +u +v + w ⎟⎟ = − + ρg z + μ ( 2 + 2 + 2 ) ∂x ∂y ∂z ⎠ ∂z ∂x ∂y ∂z ⎝ ∂t
eq2.
eq3.
eq4.
Cylindrical coordinates (r ,θ , z ) : We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity r μ =const), with a velocity field V = (u r , uθ , u z ). Incompressible continuity equation: 1 ∂ (ru r ) 1 ∂ (uθ ) ∂u z + + =0 eq a) r ∂r r ∂θ ∂z r-component: ⎛ ∂u u ∂u u2 ∂u ∂u ⎞ ρ ⎜⎜ r + u r r + θ r − θ + u z r ⎟⎟ r ∂θ r ∂r ∂z ⎠ ⎝ ∂t eq b) ⎡ 1 ∂ ⎛ ∂u r ⎞ u r ∂P 1 ∂ 2ur 2 ∂uθ ∂ 2 u r ⎤ =− + ρg r + μ ⎢ − + ⎜r ⎟− 2 + 2 ⎥ ∂r r ∂θ 2 r 2 ∂θ ∂z 2 ⎦ ⎣ r ∂r ⎝ ∂r ⎠ r
θ -component:
u ∂u uu ∂u ∂u ⎞ ⎛ ∂uθ + ur θ + θ θ + r θ + u z θ ⎟ r ∂θ r ∂r ∂z ⎠ ⎝ ∂t ⎡ 1 ∂ ⎛ ∂uθ ⎞ uθ 1 ∂P 1 ∂ 2 uθ 2 ∂u r ∂ 2 uθ ⎤ + + + ρg θ + μ ⎢ =− ⎜r ⎟− 2 + 2 ⎥ r ∂θ ∂z 2 ⎦ r ∂θ 2 r 2 ∂θ ⎣ r ∂r ⎝ ∂r ⎠ r
ρ⎜
z-component: u ∂u ∂u ∂u ⎞ ⎛ ∂u ρ⎜ z + ur z + θ z + u z z ⎟ ∂r ∂z ⎠ r ∂θ ⎝ ∂t ⎡ 1 ∂ ⎛ ∂u z ⎞ 1 ∂ 2 u z ∂ 2 u z ⎤ ∂P + ρg z + μ ⎢ + =− ⎜r ⎟+ 2 ⎥ 2 ∂z ∂z 2 ⎦ ⎣ r ∂r ⎝ ∂r ⎠ r ∂θ
eq d)
eq c)
